## An elliptical example
Just as we did with polar coordinates, we'll start with a motivating example. This is again an upside down paraboloid but now the cross-sections are ellipses. Note that the 4y^2 could be written (2y)^2 so that it is compressed in the y direction.

### Ellipse in the plane
The previous view was 3D; we can see the ellipse since we're looking at it from above. Here's the actual ellipse in the plane. This is important to understand since this is the domain of integration. It can be tricky to plot these so I've linked a tool that helps you do so.

### New coordinates
Now things get a bit more complicated because you've got to ask - where did those formulae come from? In polar coords, you've always got x=rcos(theta), y=rsin(theta) - every time. In this general CoV situation, you sometimes have to figure out a complicated function on your own.
In this "elliptic" example, I've chosen x=2scos(t), y=ssin(t) which looks a lot like polar coords but just a bit different in that the x coordinate is stretched out by the factor 2. I've also changed the variables r and theta to s and t, which are close but different because we don't want to confuse them. As we move on, I'll generally use s and t for generic coordinates.

### New coordinates (cont)
Here's another way to think about the new coordinates. The functions x and y (as functions of s and t) together map a simple rectangle in an st-plane to the xy-plane. The bounds of that rectangle will ultimately tell us the bounds of integration.

### New coordinates (cont 2)
A key aspect of polar coordinates was that they simplified, not only the domain, but also the function itself. That happens here as well. As we translate the integral from one coordinate system to the other, we again need to translate the bounds of integration, the function, and the area element dA. As before, there's a mysterious last piece that accounts for the degree to which the transformation distorts  area.


### Last piece (cont)
Here's another look at how the polar coordinate transformations distort area. Area is evenly distributed in the rectangle on the left that lives in a Cartesian r\theta-plane, when we break it up using lines of constant r and theta. The image of those lines under the transformation yields regions of various sizes. The larger r is, the larger the corresponding area. That's why we multiply by r

### Last piece (cont 2)
Our new coordinate system is a lot like polar coordinates but stretched out by an additional factor of 2. Thus, it makes sense that we might multiply by 2s.

## The Jacobian
There are lots of potential regions that we could do this with so we'd like a systematic tool to find the area distortion. The standard tool for this is called the Jacobian. Note that this involves a determinant which, as you'll learn in linear algebra, measures exactly area distortion of a linear transformation. Our transformations aren't necessarily linear but, when combined with the derivative, this gives a local measure of area distortion.

### Jacobian example
Here's the Jacobian for the general elliptic example where we stretch by the factor R1 in the x-direction and by the factor R2 in the y-direction. Note that you can play with those parameters with the parametric plotter that I mentioned earlier.

## Integrating over a parallelogram
Again, part of the power here is that you can integrate over *completely* different kinds of regions. Here, we ask - how can we integrate over a parallelogram?

### Parallelogram (cont)
Here's a specific integral to set up. Note that it is our responsibility to find the new variables that will simplify the description of the domain. I chose to set s=3x-y and t=3y-x because those expressions are constant on the sides of the parallelogram (that's exactly how the sides were defined in the previous slide). Thus, our new bounds of integration for s and t should be exactly those constants. In order to translate the integrand and to compute the Jacobian, we need x and y in terms of s and t. So we just solve that system for x and y.

### Parallelogram (cont 2)
And here's the transformation of the integral. It's not particularly hard to show that the integral evaluates to 36, but it's the setup that is the big deal.

## Plotting parametric regions
I'll again mention this tool that helps you plot parametric regions. It's not exactly perfect but it works well enough for the types of examples you might try. I've got a forum question where you'll want to use it.